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C, the Complex Numbers
Recall a basic (and I mean basic) polynomial p(x) = x2 − 1
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Recall that for p(1) = 0, p(x) must
contain at least one (x − 1) such that
p(x) = (x − 1) p1 (x)
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Next, let us consider p(x) = x2 + 1
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What we need is some value, call it i,
such that i2 = −1, leading to
p(i) = (i)2 + 1 = (−1) + 1 = 0
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Well, that problem is solved, in a sense, but i is NOT a real number, it is
classified as ‘imaginary’
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There is no reason we can’t multiply i by
real numbers, or why we can’t add it to real numbers
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The basic form of is
z = a + bi,
a is the length in the real direction and b is the length in the imaginary direction
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Everything on this plane taken together forms
the complex numbers, written C
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Doing so is actually
very easy, usually
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So, addition is just about keeping them separate:
z1 = a1 + b1 i

z 2 = a 2 + b2 i

z1 + z2 = (a1 + b1 i) + (a2 + b2 i) = (a1 + a2 ) + (b1 + b2 )i
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Next, you simply have to recall that i2 = −1
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Note that the calculation is not done until the result has been written in the
standard a + bi form
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We define
√
|z| = |a + bi| = a2 + b2
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If that is the case, then z is the hypotenuse of a right angled triangle with
sides a and bi, with lengths |a| and |b|
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There is one unique operation applied to complex numbers called the
complex conjugate
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It simply inverts the imaginary component
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The conjugate has
several uses
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Finally we get to division
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We will use the old mathematical trick: multiply
by one
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z1
z1 (z2 )
z1 z2
=
=
z2
z2 (z2 )
|z2 |2
and now we need to switch to the a and b format
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a2 + b2
a2 + b2
2
2
2
2
Again, the question isn’t answered unless it has been written in the a + bi
format
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Remember the polynomials and their factors and roots?
Well, C allows us to fully factor any polynomial
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The theorem actually covers more than the standard real valued polynomials
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The fundamental theorem may not seem like much, but we can easily
expand it
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This is easy to prove
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If it is constant, then its
order zero and we simply get p(x) = a0
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We can factor out (x − z1 ) for
p(x) = (x − z1 )p1 (x)
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One thing to note: the z1 to zn need not be unique, we may have repeats
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2
2
2

This means that the factors are 3 + i and 3 − i, or, rather, 3 + i and 3 − i
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Turns out,
it is built into ALL real polynomials
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If p(x) has a root z then p(x) must also have the root z (recall
the complex conjugate)
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In general, see what
happens if we multiply (x − z) with (x − z)
x − (a + bi) x − (a − bi) = x2 − x(a − bi) − x(a + bi) − (a + bi)(a − bi)
= x2 − x(a − bi + a + bi) − a2 − abi + abi − b2 i2
= x2 − 2ax − (a2 + b2 )
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All roots of real polynomials come in conjugate pairs
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Real numbers are unchanged by
complex conjugation, so this applies very trivially to real roots (i
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, if k ∈ R
is a root of p(x) then k = k is a root of p(x))
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First, since we have 2 + i we will also get 2 − i
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So, we divide that out of p(x), leading to
x3 − 5x2 + 9x − 5 = (x − 1)(x2 − 4x + 5) = (x − 1)(x − 2 − i)(x − 2 + i)
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5, subsections 1, 2
and 4
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bdfhj) 2
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d) 4
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b) 8
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d)
A note about 9d): By ‘irreducible’ they mean a quadratic that cannot be
factored in R

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